Page 652 - Mechatronics with Experiments

638 MECHATRONICS

R x

ϕ
y
g
i N . i
R y R y
V R Spring
x g y g

(a) (b)
FIGURE 8.19: A solenoid example. The magnetic circuit is shown on the right. The force is a
varies with motion, and the air
nonlinear function of the air gaps and current. The air gap x g
gaps y are fixed.
g
Notice that the electrical time constant of the solenoid, = L∕R, can become large
when L is large. In order to reduce the electrical time constant for fast response, one method
is to increase the effective resistance by adding external resistance in series with the coil.
As R increases, decreases. However, in order to provide the rated current at increased
resistance, the supply voltage level must be proportionally increased since i = V∕R.This
2
in turn increases the resistive heat loss, P Ri = R ⋅ i .

Example Consider the solenoid type shown in Figure 8.19. There are three air gap
paths, one x and two y spaces. The winding has N turns, and a controlled current i is
g
g
supplied to the winding. Let us determine the force generated as a function of current i
and the variable air gap x . The air gaps y on both sides are constant. Assume that the
g
g
permeability of the iron core is much larger than that of the air gaps and neglect the MMF
loss along the path in the iron. Then, the total MMF is stored in the air gaps x and y .Let
g g
A and A denote the cross-sectional areas at the air gaps.
x y
From conservation of the MMF, if we trace the flux path either on the top or bottom
section, we obtain
H ⋅ x + H ⋅ y = N ⋅ i (8.145)
x g y g
From symmetry,
Φ = 2 ⋅ Φ y (8.146)
x
B ⋅ A = 2 ⋅ B ⋅ A y (8.147)
x
y
x
Note that
B = ⋅ H x (8.148)
x
o
B = ⋅ H y (8.149)
o
y
Hence,
Φ = 2 ⋅ Φ (8.150)
x y
⋅ H ⋅ A = 2 ⋅ ⋅ H ⋅ A (8.151)
o x x o y y
This results in the following magnetic field strength relation,
A x
H = H x (8.152)
y
2A y